Question

Let W1 and W2 be subspaces of a vector space V and define
W1 + W2 := {u+v : u ∈ W1, v ∈ W2}.
Prove that span(W1 U W2) = W1 + W2.

Answer 1

Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$ and define $W_{1} + W_{2} := \{u + v : u \in W_{1}, v \in W_{2}\}$. Prove that $\text{span}(W_{1} \cup W_{2}) = W_{1} + W_{2}$.

Solution

Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$ and define $W_{1} + W_{2} := \{u + v : u \in W_{1}, v \in W_{2}\}$. In other words, $W_{1} + W_{2}$ is the collection of all vectors you can get by adding an element of $W_{1}$ to an element of $W_{2}$.

Prove that $\text{span}(W1 \cup W_2) = W_1 + W_2$ (in other words $W_1 + W_2$ is the smallest subspace containing both $W1$ and $W_2$).

To see $W_{1} + W_{2}$ is a subspace, check closure under addition and under multiplication by scalars. Let $u + v$ and $\acute{u} + \acute{v}$ be any elements of $W_{1} + W_{2}$, where $u, \acute{u} \in W_{1}$ and $v, \acute{v} \in W_{2}$, and let $a$ be any scalar. Then, since $W_{1}$ and $W_{2}$ are closed under addition and under multiplication by scalars,

$$(u + v) + (\acute{u} + \acute{v}) = (u + \acute{u}) + (v + \acute{v}) \in W_{1} + W_{2},$$ $$a(u + v) = a u + a v \in W_{1} + W_{2}.$$

Also, $W_{1} \subseteq W_{1} + W_{2}$, since every $u \in W_{1}$ can be written $u + 0 \in W_{1} + W_{2}$. For the same reason, $W_{2} \subseteq W_{1} + W_{2}$. We have shown $W_{1} + W_{2}$ is a subspace containing both $W_{1}$ and $W_{2}$.

Clearly every element $u + v \in W_1 + W_2$ is in $\text{span}(W_1 + W_2)$ so $W_1 + W_2 \subseteq \text{span}(W_1 \cup W_2)$.

To show $W_{1} + W_{2} = \text{span}(W_{1} \cup W_{2})$, it remains only to show that $\text{span}(W_{1} \cup W_{2}) \subseteq W_{1} + W_{2}$. But this must be true, because we have shown $W_{1} + W_{2}$ is a subspace containing $W_{1} \cup W_{2}$, and $\text{span}(W_{1} \cup W_{2})$ is the smallest such subspace.

So we have shown that $\text{span}(W1 \cup W_2) = W_1 + W_2$.